Answer :
Answer:
a)[tex]R = 171[/tex]μΩ
b)[tex]E = 1.7 *10^{-4} V/m[/tex]
c)[tex]R_{2} = 1.16 *10^{-4}[/tex]Ω
here * stand for multiplication
Explanation:
length of cylinder = 1.5 m
radius of cylinder = 1.1 cm
resistivity depends on the distance x from the left
[tex]p(x)=a+bx^2[/tex] ............(i)
using equation
[tex]R = \frac{pl}{a}[/tex]
let dR is the resistance of thickness dx
[tex]dR =\frac{p(x)dx}{a}[/tex]
where p(x) is resistivity l is length
a is area
[tex]\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\[/tex].........................(2)
after integration
[tex]R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}[/tex] ...............(3)
it is given [tex]p(0) = a = 2.25 * 10 ^{-8}[/tex]Ωm
[tex]p(L) = a + b(L)^2[/tex] = [tex]8.5 * 10 ^{-8}[/tex] Ωm
[tex]8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\[/tex]
(here * stand for multiplication )
on solving we get
[tex]b = 2.78* 10^{-8}[/tex] Ωm
put each value of a and b and r value in equation 3rd we get
[tex]R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}[/tex]
[tex]R = 1.71 * 10^{-4}[/tex]Ω
[tex]R = 171[/tex]μΩ
FOR (b)
for mid point x = L/2
E = p(x)L
for x = L/2
[tex]p(L/2) = a+b(L/2)^2[/tex]
for given current I = 1.75 A
so electric field
[tex]E = \frac{[a+b(L/2)^2]I }{\pi r^2}[/tex]
by substitute the values
we get;
[tex]E = 1.7 *10^{-4} V/m[/tex]
(here * stand for multiplication )
c ).
75 cm means length will be half
that is x = L/2
integrate the second equation with upper limit L/2
Let resistance is [tex]R_{1}[/tex]
so after integration we get
[tex]R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}[/tex]
substitute the value of a , b and L we get
[tex]R_{1} = 5.47 * 10 ^{-5}[/tex]Ω
for second half resistance
[tex]R_{2} = R- R_{1}[/tex]
[tex]R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}[/tex]
[tex]R_{2} = 1.16 *10^{-4}[/tex]Ω
(here * stand for multiplication )
